Find the extremal to the functional and discuss whether they provide a max/min

Nov 2006
17
1
I am having a hard time getting my head around Functionals and Calculus of Variations,

My question is: Given a functional and using the Euler-Lagrange equation to find an extremal, how do we show that the extremal provides a min/max (if it does)

The question I am working on is

\(\displaystyle J(y) = \int_{0}^{1} ((y')^2 -y)dx\) with \(\displaystyle y(0)=0, y(1)=1\)

I found the extremal to be: \(\displaystyle y(x) = \frac{-1}{4}x^2 +\frac{5}{4}x\) which I am told is a minimum to the functional problem.

However I am unsure on what is sufficient to show this, in the notes I have it is shown that:

\(\displaystyle J(y+f) = J(y) + \int_{0}^{1}(f')^2dx \geq J(y)\) where f is continuously differentiable on the interval 0,1 with \(\displaystyle y(0)=y(1)=0\)

Thanks in advance!
 
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GJA

Jul 2012
109
32
USA
Hi dtwazere.

Calculus of Variations can seem a little daunting, however hopefully a few comments will make things seem a little less scary.

1) You can think of the term "functional" in this case as something that sends functions to real numbers. The definition of a functional is actually a little more technical than that (it is a map from a vector space to its underlying scalar field - see Functional (mathematics) - Wikipedia, the free encyclopedia) - in your case the vector space is the space of continuously differentiable functions with \(\displaystyle y(0)=0\) and \(\displaystyle y(1)=1\) and the "field" is the real numbers. However, thinking of a functional as something whose domain is a certain set of functions (in your case continuously differentiable functions with \(\displaystyle y(0)=0\) and \(\displaystyle y(1)=1\)) and whose range is the real numbers is good enough for now.

2) You have actually done most of the hard work already by solving the Euler-Lagrange equation to find a possible minimizer. I will denote your possible minimizer by \(\displaystyle y_{min};\) i.e. \(\displaystyle y_{min}(x)=-\frac{1}{4}x^{2}+\frac{5}{4}x.\) Note that my notation is a little dangerous because we haven't actually proved that \(\displaystyle y_{min}\) is actually a minimizer of \(\displaystyle J\) yet. The last thing we need to note is that the condition \(\displaystyle f(0)=0=f(1)\) was left out in the original post where you mentioned \(\displaystyle J(y+f).\) The condition \(\displaystyle f(0)=1=f(1)\) is essential; \(\displaystyle y_{min}+f\) represents a small pertubation of \(\displaystyle y_{min}\), however we don't perturb the endpoints of \(\displaystyle y_{min}\), which is where the condition \(\displaystyle f(0)=1=f(1)\) comes from.

Now we compute

\(\displaystyle J(y_{min}+f)=\int_{0}^{1}[(y_{min}+f)']^{2}-(y_{min}+f)dx\)

expanding this we get

\(\displaystyle J(y_{min}+f)=\int_{0}^{1}(y_{min}')^{2}-y_{min}dx+\int_{0}^{1}2y_{min}'f'dx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx\)

Which becomes

\(\displaystyle J(y_{min}+f)=J(y_{min})+\int_{0}^{1}2y_{min}'f'dx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx\)

If we integrate by parts on the second term on the RHS we obtain
\(\displaystyle 2\int_{0}^{1}y_{min}'f'dx=2(y'f |_{0}^{1} -\int_{0}^{1}y_{min}''fdx)\)

Now use \(\displaystyle f(0)=0=f(1)\) and \(\displaystyle y_{min}''=-\frac{1}{2}\) on the previous line to get

\(\displaystyle 2\int_{0}^{1}y_{min}'f'dx=\int_{0}^{1}fdx\)

Using the last line in \(\displaystyle J(y_{min}+f)\) we have

\(\displaystyle J(y_{min}+f)=J(y_{min})+\int_{0}^{1}fdx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx=J(y_{min})+\int_{0}^{1}(f')^{2}dx\geq J(y_{min})\)

This proves that \(\displaystyle y_{min}\) really is a minimizer of \(\displaystyle J\) for the class of functions that you're considering.

Does this help? Let me know if anything is unclear.

Good luck!
 
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Nov 2006
17
1
This really helps, I think I understand most of what you put, thanks a lot!